Question: Evaluate the iterated integral. $ \int_1^3 \left( \int_{-1}^1 4x^3 - 3xy^2 - 1 \, dx \right) dy =$ Choose 1 answer: Choose 1 answer: (Choice A) A $84$ (Choice B) B $128$ (Choice C) C $-4$ (Choice D) D $15$
Solution: Evaluate the inner integral: $\begin{aligned} \int_1^3 \left( \int_{-1}^1 4x^3 - 3xy^2 - 1 \, dx \right) dy &= \int_1^3 \left[ x^4 - \dfrac{3x^2y^2}{2} - x \right]_{-1}^1 dy \\ \\ &= \int_1^3 -2 \, dx \end{aligned}$ Evaluate the outer integral: $\begin{aligned} \int_1^3 -2 \, dx &= -2x \bigg|_1^3 = -4 \end{aligned}$ The answer: $ \int_1^3 \left( \int_{-1}^1 4x^3 - 3xy^2 - 1 \, dx \right) dy = -4$